Chapter 11.2, Problem 10CYU

To determine

Find the sum of the arithmetic series where $a_1=145;d=5,n=21$

Answer to Problem 10CYU

Sum = 4095

Explanation of Solution

Given:

Given: $a_1=145;d=5,n=21$

Concept Used:

The general formula for sum of arithmetic series is: ${\text{S}}_{\text{n}}=\frac{n}{2}[2a_1+(n-1)·d]$

If nth term is given, then the formula: ${\text{S}}_{\text{n}}=\frac{n}{2}(a_1+a_n)$

Calculation:

  $a_1=145;d=5,n=21;\text{find}{\text{S}}_{\text{n}}$

Step 1: Find $a_n$

Step 2: Find ${\text{S}}_{\text{n}}$

  $\begin{array}{l}{a}_1=145;d=5,n=21;\text{find}{a}_{\text{n}}\\ a_n=a_1+(n-1)·d\\ a_n=145+(21-1)·5\\ a_n=145+100\\ a_n=245\end{array}$

  $\begin{array}{l}{a}_1=145;a_n=245,n=21;\text{find}{\text{S}}_{\text{n}}\\ {\text{S}}_{\text{n}}=\frac{n}{2}(a_1+a_n)\\ {\text{S}}_{\text{n}}=\frac{21}{2}(145+245)\\ {\text{S}}_{\text{n}}=21·195=4095\\ {\text{S}}_{\text{n}}=4095\end{array}$

Thus, the sum of the series is 4095.